「 SPOJ GSS3 」 Can you answer these queries III

# 題目大意

GSS3 - Can you answer these queries III

須要你維護一種數據結構，支持兩種操做：

• 單點修改
• 求一個區間的最大子段和

 

# 解題思路

一個區間的最大子段和(GSS)，只能經過三種方式轉移而來。

• 左兒子的最大子段和
• 右兒子的最大子段和
• 左兒子的最大右子段和+右兒子的最大左子段和

那這就比較好辦了。只須要維護四個東西就能夠了

• sum，區間和
• gss，最大子段和
• gssl，最大左子段和
• gssr，最大右子段和

emmm，比較可作。

 

# 代碼

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
inline int read() {
    int x = 0, f = 1; char c = getchar();
    while (c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while (c <= '9' && c >= '0') {x = x*10 + c-'0'; c = getchar();}
    return x * f;
}
const int maxn = 5e4+3, inf = 2147483647;
int n, m, opt, l, r;
struct node {
    int l, r, gss, gssr, gssl, sum;
}tree[maxn << 2];
struct TREE {
    #define Lson (k << 1)
    #define Rson ((k << 1) + 1)
    inline int MAX(int a, int b, int c) {
        return max(max(a, b), c);
    }
    inline void build(int k, int ll, int rr) {
        tree[k].l = ll, tree[k].r = rr;
        if(tree[k].l == tree[k].r) {
            tree[k].sum = read();
            tree[k].gss = tree[k].gssr = tree[k].gssl = tree[k].sum;
            return ;
        }
        int mid = (tree[k].l + tree[k].r) >> 1;
        build (Lson, tree[k].l, mid);
        build (Rson, mid+1, tree[k].r);
        tree[k].sum = tree[Lson].sum + tree[Rson].sum;
        tree[k].gss = MAX(tree[Lson].gss, tree[Rson].gss, tree[Lson].gssr+tree[Rson].gssl);
        tree[k].gssr = max(tree[Rson].gssr, tree[Rson].sum+tree[Lson].gssr);
        tree[k].gssl = max(tree[Lson].gssl, tree[Lson].sum+tree[Rson].gssl);
    }
    inline void update(int k, int pos, int num) {
        if(tree[k].l == tree[k].r && tree[k].l == pos) {
            tree[k].sum = num;
            tree[k].gss = tree[k].gssr = tree[k].gssl = tree[k].sum;
            return ;
        }
        int mid = (tree[k].l + tree[k].r) >> 1;
        if(pos <= mid) update(Lson, pos, num);
        else update(Rson, pos, num);
        tree[k].sum = tree[Lson].sum + tree[Rson].sum;
        tree[k].gss = MAX(tree[Lson].gss, tree[Rson].gss, tree[Lson].gssr+tree[Rson].gssl);
        tree[k].gssr = max(tree[Rson].gssr, tree[Rson].sum+tree[Lson].gssr);
        tree[k].gssl = max(tree[Lson].gssl, tree[Lson].sum+tree[Rson].gssl);
    }
    inline node query(int k, int L, int R) {
        if(tree[k].l == L && tree[k].r == R) return tree[k];
        int mid = (tree[k].l + tree[k].r) >> 1;
        if(L > mid) return query(Rson, L, R);
        else if(R <= mid) return query(Lson, L, R);
        else {
            node lson, rson, res;
            lson = query(Lson, L, mid);
            rson = query(Rson, mid+1, R);
            res.sum = lson.sum + rson.sum;
            res.gss = MAX(lson.gss, rson.gss, lson.gssr+rson.gssl);
            res.gssl = max(lson.gssl, lson.sum+rson.gssl);
            res.gssr = max(rson.gssr, rson.sum+lson.gssr);
            return res;
        }
    }
}T;
int main() {
    n = read(), T.build(1, 1, n);
    m = read();
    for(int i=1; i<=m; i++) {
        opt = read(), l = read(), r = read();
        if(opt == 1) printf("%d\n", T.query(1, l, r).gss);
        else T.update(1, l, r);
    }
}